Second differential of the norm in an infinite dimensional Hilbert space Let $f: E \to \mathbb{R}$ sending  $x \mapsto \|x\|$ and make some simple hypothesis


*

*$E$ is a Hilbert Space

*Let's say that the norm $\|\cdot\|$ is derived from a scalar product


*

*[solved] So we can easily find the différential: $D\|\cdot\|(x)(h)=\langle x/\|x\|,h\rangle$ with $\nabla{ \|\cdot\|}(x)=x/\|x\| \, (grad)$ as shown below:
http://www.les-mathematiques.net/phorum/file.php?4,file=43281
But computing the second order differential seems really complicated. I don't even know how to proceed! 

*What about the 3rd order ? $D^3f(x)$ ?  
 A: So, you want to differentiate the map $T(X) = x/\|x\| $. For any vector $h$, the derivative of $T(x+th)$ with respect to $t$ at $t=0$ can be computed using the ordinary quotient rule: 
$$\frac{d}{dt}_{| t=0} \frac{x+th}{\|x+th\|}
= \frac{h\|x\| - x\langle x,h\rangle/\|x\| }{\|x \|^2 }$$
So, the derivative at $x$ is the linear operator
$$
h\mapsto \frac{h}{\|x\|}- \frac{x\langle x,h\rangle}{\|x\|^3}
$$
which is the projection of $h$ onto the orthogonal complement of $x$, multiplied by  $\|x\|^{-1}$.
A: I tried an other way: by composition and multiplication rules. Let's call $g:=1/\|\bullet \|$.


*

*$Dg(x).(h)=-\frac 1 {||x||^3}<x,h>$ (because $g=(1/\bullet) \circ f$ so $
  Dg(x)=-1/f(x)^2.Df(x) $)

*$T=Id . g$ (so using $D(fg)_{a}(h)=Df_{a}(h)g(a)+f(a)Dg_{a}(h)$) we have  $DId(x)(h)=h$ (linear application get the same derivative).
finally $DT(x)(h)=h.1/\|x \|+x.-\frac 1 {||x||^3}<x,h>$.


Finally $D(\nabla{ \|\cdot\|})(x)(h)=\frac{h}{\|x\|}- \frac{x\langle x,h\rangle}{\|x\|^3}$ but we don't have  $D^2( \|\cdot\|)(x)$!
We know that:  $D( \|\cdot\|)(x)=<\nabla{( \|\cdot\|})(x),\cdot>$. So if we call $p:a \to <a,\cdot>$ we have $D( \|\cdot\|)(x)=(p \circ \nabla{( \|\cdot\|}))(x)$ with p a linear function. Finally  $D^2( \|\cdot\|)(x)(h):=D(D( \|\cdot\|))(x)(h)=p \circ (D\nabla{( \|\cdot\|}))(x)(h)=<\frac{h}{\|x\|}- \frac{x\langle x,h\rangle}{\|x\|^3},\cdot>$. 
$D^2( \|\cdot\|)(x)(h)(k)=<\frac{h}{\|x\|}- \frac{x\langle x,h\rangle}{\|x\|^3},k>$
Remark: The funny thing (i just discovered it!) is that: $\nabla({Df})(x)=D(\nabla({Df}))$. $D(Df)(x)(h):=<\nabla({Df})(x),h>=<k/\|x\|-x<x,k>/\|x\|^3,h>$ (rearranging the term to move the h on the other side)
